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3 and 4 .Determinants and Matrices
medium
The value of $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$is
A
$1$
B
$0$
C
$(a - b)(b - c)(c - a)$
D
$(a + b)(b + c)(c + a)$
Solution
(c) $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$=$\left| {\,\begin{array}{*{20}{c}}0&0&1\\{c(b – a)}&{a(c – b)}&{ab}\\{b – a}&{c + a}&{a + b}\end{array}\,} \right|$
$\{ {C_1} \to {C_1} – {C_2},\,{C_2} \to {C_2} – {C_3}\} $
= $(b – a)\,\,(c – b)\,\,\left| {\,\begin{array}{*{20}{c}}0&0&1\\c&a&{ab}\\1&1&{a + b}\end{array}\,} \right|$= $(b – a)\,(c – a)\,\,(c – a)$
$ = (a – b)\,\,(b – c)\,\,(c – a)$.
Standard 12
Mathematics